# Math Help - method of difference

1. ## method of difference

can anyone please explain how does the formula work? i cant understand . if i sub r=1 into the first eq, i would get f(1)-f(1+1) , not f(1)-f(n+1), when should i use the first and when should i use the second formula, can you givr relevant examples please? much appreciated!

2. ## Re: method of difference

$\sum_{r=1}^n=(f(1)-f(2))+(f(2)-f(3))+(f(3)-f(4))+\dots (f(n-1)-f(n))+(f(n)-f(n+1))=$

$\sum_{r=1}^n=f(1)+(-f(2)+f(2))+(-f(3)+f(3))+(-f(4)+f(4))+\dots (-f(n))+f(n))-f(n+1)=$

$f(1)-f(n+1)$

similar for the other formulas.